Does this liminf characterization hold true?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $$0<a:=\liminf_{|t|\to\infty} tf(t) <\infty.$$ I am interested in writing the above relation by using the definition. On by lecture notes, I have written that it corresponds to say that, fixed $\varepsilon>0$, it is $$ tf(t)\geq(a-\varepsilon) - c_1 \quad\mbox{ for all } t, $$ where $c_1$ denotes a positive constant. Now, I would like to obtain a similar relation but using the "reverse" inequality, I mean, it is true that $$ tf(t)\leq (a+\varepsilon) +c_2\quad\mbox{ for all } t $$ holds too?

Thank you in advance!

Answer 1

Let $g(t)=\inf \{x f(x): |x|\ge t\}$. This is an increasing function with value in ${\mathbb R}\cup \{-\infty\}$ and we have $\lim_{t\to\infty} g(t) = a$. The condition that $0<a<+\infty$ is equivalent to the fact that $0<g(t_0)$ for some $t_0$ and $g$ is bounded from above by some constant $A$. This entails the two conditions which characterize the fact that $a \in (0, \infty)$

• $\exists m > 0,\exists t_0 \in {\mathbb R}, \forall t, |t|\ge t_0 \Rightarrow t f(t)\ge m$
• $\exists A\in{\mathbb R}, \exists t_n, |t_n|\to \infty, t_n f(t_n)\le A$